What is the instantaneous rate of change of f(x)=x−x+7 at x=0?

"What is the instantaneous rate of change of..." was answered by our experts

Kienastsrx

Answered question

2022-03-16

What is the instantaneous rate of change of

f (x) = \frac{x}{- x + 7}

at x=0?

Answer & Explanation

nida0694ii5

Beginner2022-03-17Added 6 answers

The instantaneous rate of change is the first derivative calculated at
x=0
Hence

f^{'} (x) = \frac{7}{{(7 - x)}^{2}}

f^{'} (0) = \frac{7}{7^{2}} = \frac{1}{7}

Talaminiu2d

Beginner2022-03-18Added 4 answers

Apply the quotient rule, which is:

\frac{d}{d x} \frac{f (x)}{g (x)} = \frac{- f (x) \frac{d}{d x} g (x) + g (x) \frac{d}{d x} f (x)}{g^{2} (x)}

f(x)=x and g(x)=7-x
To find

\frac{d}{d x} f (x) :

Apply the power rule: x goes to 1
To find

\frac{d}{d x} g (x) :

Differentiate 7-x term by term:
The derivative of the constant 7 is zero.
The derivative of a constant times a function is the constant times the derivative of the function.
Apply the power rule: x goes to 1
So, the result is: -1
The result is: -1
Now plug in to the quotient rule:

\frac{7}{{(7 - x)}^{2}}

Now simplify:

\frac{7}{{(x - 7)}^{2}}

If we putting the value of x, we get:
0

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